Question

我的代码允许我执行相当复杂的任务（至少对我来说）：

import csv
import os.path
#open files + readlines
with open("C:/Users/Ivan Wong/Desktop/Placement/Lists of targets/Mouse/UCSC to Ensembl.csv", "r") as f:
    reader = csv.reader(f, delimiter = ',')
    #find files with the name in 1st row
    for row in reader:
        graph_filename = os.path.join("C:/Python27/Scripts/My scripts/Top targets",row[0]+"_nt_counts.txt.png")
        if os.path.exists(graph_filename):
            y = row[0]+'_nt_counts.txt'  
            r = open('C:/Users/Ivan Wong/Desktop/Placement/fp_mesc_nochx/'+y, 'r')
            k = r.readlines()
            r.close
            del k[:1]
            k = map(lambda s: s.strip(), k)
            interger = map(int, k)   
            import itertools
            #adding the numbers for every 3 rows
            def grouper(n, iterable, fillvalue=None):
                "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
                args = [iter(iterable)] * n
                return itertools.izip_longest(*args, fillvalue=fillvalue)
            result = map(sum, grouper(3, interger, 0))       
            e = row[0]
            print e
            cDNA = open('C:/Users/Ivan Wong/Desktop/Placement/Downloaded seq/Mouse/MOUSE_mRNAs.txt', 'r')
            seq = cDNA.readlines()
            # get all lines that have a gene name
            lineNum = 0;
            lineGenes = []
            for line in seq:
                lineNum = lineNum +1
                if '>' in line:
                    lineGenes.append(str(lineNum))
                if '>'+e in line:
                    lineBegin = lineNum

            cDNA.close

            # which gene is this
            index1 = lineGenes.index(str(lineBegin))
            lineEnd = lineGenes[index1+1]           
# linebegin and lineEnd now give you, where to look for your sequence, all that 
# you have to do is to read the lines between lineBegin and lineEnd in the file
# and make it into a single string.            
            lineEnd = lineGenes[index1+1]
            Lastline = int(lineEnd) -1

# in your code you have already made a list with all the lines (q), first delete
# \n and other symbols, then combine all lines into a big string of nucleotides (like this)     
            qq = seq[lineBegin:Lastline]
            qq = map(lambda s: s.strip(), qq)
            string  = ''
            for i in range(len(qq)):
                string = string + qq[i]
# now you want to get a list of triplets, again you can use the for loop:
# first get the length of the string
            lenString = len(string);
# this is your list codons
            listCodon = []
            for i in range(0,lenString/3): 
                listCodon.append(string[0+i*3:3+i*3])
            proper_result = '\n'.join('%s, %s' % (nr, codon) for nr, codon in zip(result, listCodon))
            with open(e+'.csv','wb') as outfile:
                outfile.writelines(proper_result)

这些代码从.csv中读取文件，从具有相同名称的文件的文件夹中识别，如果存在，则继续处理某些数据并将其写入.csv 与他们一起，我的outfiles现在看起来像 outfile

它看起来完全没问题，但有一个问题，我从我的数据（我用不同的方式检查过）得知第二列应该比我的更长。我认为这是因为当BOTH结果（数字）和listCodon（字母）存在时，代码正在写文件，因此我遗漏了一些东西。我该如何解决？

我试着在文件写入之前打印listCodon，发现所有三元组仍然存在，所以我猜这个问题就在这里：

proper_result = '\n'.join('%s, %s' % (nr, codon) for nr, codon in zip(result, listCodon))

Answer 1

zip会在其任意的任何停止时停止（因为否则它将无法知道填充空白的内容！）：

返回的列表的长度被截断为最短参数序列的长度。

如果要将较短的可迭代填充到最长的长度，请使用izip_longest（它将可选参数用作填充值）。

Python outfile没有写出一切

1 个答案: